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Suppose $a$ is a point of the metric space $S$. Define $g(p) = d(a,p)$ with $p \in S$. Prove $g$ is uniformly continuous.

Also, if possible, don't use Lipschitz continuity or denseness. We haven't covered those in class so I'll have no idea what you're talking about.

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$|g(p)-g(q)|\leqslant d(p,q)$. –  Did Dec 13 '11 at 20:34
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$g(p)+d(p,q)\geq g(q)$ by the triangle inequality –  yoyo Dec 13 '11 at 20:38
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You want to prove that for all $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $d(p,q)\lt \delta$, then $|g(p)-g(q)|\lt\epsilon$.

By the triangle inequality, $$d(p,a) \leq d(p,q)+d(q,a),$$ so $$d(p,a)-d(q,a) \leq d(p,q).$$ Symmetrically, $$d(q,a) \leq d(q,p) + d(p,a),\qquad\text{so}\qquad d(q,a)-d(p,a)\leq d(p,q).$$ Therefore, $$|d(q,a)-d(p,q)|\leq d(p,q).$$

Can you finish this off?

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